Physics100ACh4Guide - CHAPTER 4 Giancoli Physics Study...

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CHAPTER 4 Giancoli: Physics Study GuideDr. LeeSummary:Newton’s Laws:1. If there is no net force, an object will be at rest or at constant velocity.2. F = ma (F is net force)3. Action is equal and opposite to reactionForce on an object exerted by the earth = its weight = mgFriction:1. Static: max friction = μSFN2. Kinetic: μKFNSolution of some problems.(Note: capital X means multiplication)Isolate the mass that is under consideration and consider all forces acting on it.2.If we select the bike and rider as the object, we apply Newton’s second law to find the mass:F = ma; 225 N = m(2.20 m/s2), which gives m= 116 kg.4.Without friction, the only horizontal force is the tension. We apply Newton’s second law to the car:F = ma; FT= (1050 kg)(1.20 m/s2) = 1.26×103N.7.The acceleration can be found from the car’s one-dimensional motion:v= v0+ at; 0 = [(90 km/h)/(3.6 ks/h)] + a(8.0 s), which gives a= – 3.13 m/s2.We apply Newton’s second law to find the required average forceF = ma; F= (1100 kg)(– 3.13 m/s2) = – 3.4×103N.The negative sign indicates that the force is opposite to the velocity.12.We write ∑F= ma from the force diagram for the car:y-component: FTmg= ma, or FT= m(a+ g) = (1200 kg)(0.80 m/s2+ 9.80 m/s2) = 1.3×104N.17.The maximum tension will be exerted by the motor when the elevator has the maximum acceleration. We write ∑F= ma from the force diagram for the elevator:1mgFTymgFTy
y-component: FTmaxmg= mamax ; 21,750 N – (2100 kg)(9.80 m/s2) = (2100 kg)amax , which gives amax= 0.557 m/s2.27. (a)For the components of the net force we haveFax= – F1= – 10.2 N;Fay= – F2= – 16.0 N.We find the magnitude fromFa2= Fax2+ Fay2= (– 10.2 N)2+ (– 16.0 N)2, which gives Fa= 19.0 N.We find the direction fromtan α= |Fay |/|Fax |= (16.0 N)/(10.2 N) = 1.57, which gives α= 57.5° below – x-axis.The acceleration will be in the direction of the net force:aa= Fa/m= (19.0 N)/(27.0 kg) = 0.702 m/s2, 57.5° below – x-axis.

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